Applying the implicit function theorem to a system of equations

[Mina Farag]

Homework Statement

Prove that the following system:

F(x,y,z)=2*x^2+y^2+1−z^2=0
G(x,y,z)=2*x^2+2*y^2−z^2=0

can be solved for z and y as functions of x.

Furthermore, provide the values of the points that allow the parameterization.

Relevant Equations

Jacobian of a system, implicit function theorem, Cramer's rule.

My attempt:

According to the implicit function theorem as long as the determinant of the jacobian given by ∂(F,G)/∂(y,z) is not equal to 0, the parametrization is possible.

∂(F,G)/∂(y,z)=4yzMeaning that all points where z and y are not equal to 0 are possible parametrizations.

My friend's solution:

Same as previous, conclude that according to the implicit function theorem it is all points where z and y are not 0. However, test the possibilities where y=0and z=0

• y=0: by subtracting F−Gyou get 1=0, which is impossible.
• z=0: gives G=2*x^2+2*y^2=0 which gives x=y=0, which is impossible for F, as it gives 1=0.

My question:

Is it necessary to consider the possibilities of y=0 and z=0. These possibilities should be impossible as either one makes the determinant of the jacobian equal 0 which if you were to use Cramer's rule to find the partial derivative: ∂(y)/∂(x) would lead to dividing by zero.

Furthermore, it leads to the same conclusion. Is there a reason to why it is important and are there cases where it would not lead to the same conclusion?

 

[Gaussian97]

As far as I remember the implicit theorem gives only a sufficient condition, but not a necessary one. Therefore, you cannot say that for $y,z=0$ it's impossible unless you check explicitly by other methods.

 

[mathwonk]

I could be wrong, as I am only visualizing this in my head, not checking sources, but whether it is a necessary condition seems to depend on whether you also want the dependence of z,y on x to be differentiable. If so, it seems to be a matter of whether the set defined by the two equations is a smooth curve that is (locally) a graph of a differentiable function (z(x), y(x)). This would seem to require that the linear approximation R^3-->R^2 to the map (F(x,y,z), G(x,y,z)) have rank two, and have a kernel that projects isomorphically onto the x axis. This would seem to mean the kernel is not orthogonal to the x axis, hence not contained in the y,z plane, hence that the given 2x2 jacobian matrix ∂(F,G)/∂yz be invertible, i.e. that the jacobian determinant be non zero.

E.g. in lower dimensions, it is true that given z = F(x,y) = x - y^3, one can solve x-y^3 = 0 for y in terms of x, via y = x^1/3, even near (0,0) where ∂F/∂y = -3y^2 = 0, but y(x) = x^1/3 is not differentiable there. I.e. the curve F=0 is smooth at (0,0) but the tangent line is orthogonal to the x axis.

Does this seem more or less true?

 

[Mina Farag]

I could be wrong, as I am only visualizing this in my head, not checking sources, but whether it is a necessary condition seems to depend on whether you also want the dependence of z,y on x to be differentiable. If so, it seems to be a matter of whether the set defined by the two equations is a smooth curve that is (locally) a graph of a differentiable function (z(x), y(x)). This would seem to require that the linear approximation R^3-->R^2 to the map (F(x,y,z), G(x,y,z)) have rank two, and have a kernel that projects isomorphically onto the x axis. This would seem to mean the kernel is not orthogonal to the x axis, hence not contained in the y,z plane, hence that the given 2x2 jacobian matrix ∂(F,G)/∂yz be invertible, i.e. that the jacobian determinant be non zero.

E.g. in lower dimensions, it is true that given z = F(x,y) = x - y^3, one can solve x-y^3 = 0 for y in terms of x, via y = x^1/3, even near (0,0) where ∂F/∂y = -3y^2 = 0, but y(x) = x^1/3 is not differentiable there. I.e. the curve F=0 is smooth at (0,0) but the tangent line is orthogonal to the x axis.

Does this seem more or less true?

Yes, this is quite accurate. In order for a function y(x) in F(x,y(x))=0 to exist, we must be able to find its derivative and hence there exists a nonvertical tangent line near points where that is possible.
For a system of equations that implies we can find the partial derivatives, which do exist iff the determinant of the jacobian is not zero and hence the matrix is invertible (the matrix is the one corresponding to the values of the elements in the vector [ ∂(y)/∂(x), ∂(z)/∂(x)]). This is the is more or less a consequence of Cramer's rule. If you were to find the value of ∂(y)/∂(x) and ∂(z)/∂(x) by implicitly differentiating F and G in, the values would be given by ∂(y)/∂(x)=(∂(F,G)/∂(x,z))/(∂(F,G)/∂(y,z)), ∂(z)/∂(x)=(∂(F,G)/∂(x,z))/(∂(F,G)/∂(y,z)). As you can see the denominator is the same for both and is given by the determinant of the jacobian to (∂(F,G)/∂(y,z), which is 4yz. Hence, as long as y and z are not 0 the denominator is not zero and we have an invertible matrix and can solve ∂(y)/∂(x) and ∂(z)/∂(x). This begs the question of whether it is necessary or not to consider y=0 or z=0, or whether it is even possible to do so.

 

[Gaussian97]

Look at the simple example of
$$F(x,y,z)=x-z=0$$
$$G(x,y,z)=y^3-x-z=0$$
The jacobian is
$$
\left|\frac{\partial(F,G)}{\partial(y,z)}\right|=\begin{vmatrix}
0 & -1 \\
3y^2 & -1 \\
\end{vmatrix} = 3y^2
$$
And obviously you can write $y, z$ as functions of $x$
$$y(x)=\sqrt[3]{2x},\qquad z(x)=x$$
whithout any problem at $y=0$

 

[mathwonk]

In this example, when y=0, then x = z =0 also. I.e. you are looking at a point where both x = z and x = -z, hence this occurs only at the point (0,0,0). Thus near this point you are solving y^3 = 2x, with x near zero. Hence, as you correctly said, you have the function y(x) = (2x)^1/3, which is however not differentiable near x= 0.

This seems then to be exactly the problem indicated earlier. What do you think?

 

[Gaussian97]

In this example, when y=0, then x = z =0 also. I.e. you are looking at a point where both x = z and x = -z, hence this occurs only at the point (0,0,0). Thus near this point you are solving y^3 = 2x, with x near zero. Hence, as you correctly said, you have the function y(x) = (2x)^1/3, which is however not differentiable near x= 0. This seems then to be exactly the problem indicated earlier.

Yes, but in the original post, there is nothing about derivatives. My example proves that even if the conditions of the IFT doesn't exist, you can still have a function that satisfies the equations. So IFT tells you what are the points where you can define the functions $y$ and $z$, but you still need to check for all the other points manually.

 

[mathwonk]

I agree completely. Perhaps you did not read my post #3 on this matter of differentiability implying necessity, where I also gave essentially the same example as you did, in the absence of differentiability.

"...whether it is a necessary condition seems to depend on whether you also want the dependence of z,y on x to be differentiable."

"..it is true that given z = F(x,y) = x - y^3, one can solve x-y^3 = 0 for y in terms of x, via y = x^1/3, even near (0,0) where ∂F/∂y = -3y^2 = 0, but y(x) = x^1/3 is not differentiable there."

so i think we agree. cheers!

 

[Gaussian97]

Yes, I totally missed up, sorry

 

[mathwonk]

actually you are quite correct!

 

[mathwonk]

If I may be forgiven for derailing the thread by a personal comment, as a student in the 1960's when abstraction was the rage, I "learned" the implicit function theorem somewhat as follows: If F:V-->W is a smooth map of Banach spaces, with F(0) = 0, and if the Frechet derivative of F at 0 is surjective, with kernel that splits as a direct summand of V, then in some nbhds of 0 in V and in W, and after a smooth change of coordinates, F becomes a projection ker(dF(0)) + W --> W.

Then when someone remarked that the implicit function theorem gives a (sufficient) criterion for when one can use a system of equations F1(x1,...,xn) = F2(x1,...,xn) = ... = Fr(x1,...,xn) = 0, to solve for some of the variables xj in terms of the others, I did not know what they were talking about!